Bochner Representable Operators on Spaces of Continuous Functions

Abstract

Let $C_b(X)$ be the Banach lattice of all bounded continuous real-valued functions on a completely regular Hausdorff space X and $\beta$ denote the natural strict topology on $C_b(X)$. For a Banach space $(E,\Vert \cdot \Vert _E)$, a linear operator $T:C_b(X)\rightarrow E$ is said to be tight if $\Vert T(u_\alpha )\Vert _E\rightarrow 0$ whenever $(u_\alpha )$ is a uniformly bounded net in $C_b(X)$ such that $u_\alpha \rightarrow 0$ uniformly on all compact sets in X. It is shown that a linear operator $T:C_b(X)\rightarrow E$ is nuclear tight if and only if T is a nuclear operator between the locally convex space $(C_b(X),\beta )$ and a Banach space E and if and only if T is Bochner representable, that is, there exist a positive Radon measure $\mu$ on X and a E-valued $\mu$-Bochner integrable function g on X so that $T(u)=\int _X u(x)g(x)d\mu$ for all $u\in C_b(X)$.

1 Introduction and Preliminaries

Nuclear operators on the Banach space C(X) of continuous functions on a compact Hausdorff space X have been studied intensively (see [5, 8, 11, 16,17,18, 23]). In particular, due to Tong [23, Theorem 1.2] a linear operator T from C(X) to a Banach space E is nuclear if and only if T is Bochner representable (see also [5, Theorem 4, pp. 173–174], [18, Proposition 5.30]).

The main aim of the present paper is to extend and generalize this study to the setting of X being a completely regular Hausdorff space (see Theorems 2.2, 3.2 and Corollary 3.3).

Throughout the paper we assume that X is a completely regular Hausdorff space. Let ${\mathcal B}o$ (resp. ${\mathcal B}$) denote the $\sigma$-algebra of all Borel sets (resp., the algebra of all Baire sets) in X.

Let $C_b(X)$ (resp. $B({\mathcal B}o)$) denote the Banach lattice of all bounded continuous (resp. bounded ${\mathcal B}o$-measurable) real-valued functions on X, equipped with the uniform norm $\Vert \cdot \Vert _\infty$. Let $C_b(X)'$ and $C_b(X)''$ stand for the dual and the bidual of $C_b(X)$, respectively.

Recall that a countably additive real-valued measure $\lambda$ on ${\mathcal B}o$ is called a Radon measure if for every $A\in {\mathcal B}o$ and $\varepsilon >0$, there exist a compact set K and an open set O in X with $K\subset A\subset O$ such that $|\lambda |(O\smallsetminus K)\le \varepsilon$

By $rca({\mathcal B}o)$ we denote the Banach lattice of all real-valued Radon measures $\lambda$, equipped with the norm $\Vert \lambda \Vert :=|\lambda |(X)$.

From now on we assume that $(E,\Vert \cdot \Vert _E)$ is a real Banach space.

For $\mu \in rca({\mathcal B}o)^+$, let $L^1(\mu ,E)$ denote the Banach space of $\mu$-equivalence classes of all E-valued Bochner integrable functions g on X, equipped with the norm $\Vert g\Vert _1:=\int _X\Vert g(x)\Vert _E\,d\mu$.

Following [23] we say that a linear operator $T:C_b(X)\rightarrow E$ is Bochner representable if there exist a measure $\mu \in rca({\mathcal B}o)^+$ and a function $g\in L^1(\mu ,E)$ so that

$$\begin{aligned} T(u)=\int _X u(x)\,g(x)\,d\mu \ \ \text{ for } \text{ all } \ \ u\in C_b(X). \end{aligned}$$

Recall that a linear operator $T:C_b(X)\rightarrow E$ is said to be tight if $\Vert T(u_\alpha )\Vert _E\rightarrow 0$ whenever $(u_\alpha )$ is a uniformly bounded net in $C_b(X)$ such that $u_\alpha \rightarrow 0$ uniformly on all compact sets in X (see [15, Definition 1.1]). Note that if X is compact, then every bounded operator $T:C_b(X)\rightarrow E$ is tight.

The concept of nuclear operators between Banach spaces is due to Grothendieck [9, 10] (see also [26, p. 279], [16, Chap. 3], [17, 5, Chap. 6], [8, Chap. 5], [18, 22, 24]).

Recall (see [26, p. 279], [22]) that a linear operator $T:C_b(X)\rightarrow E$ is said to be nuclear if there exist a bounded sequence $(\Psi _n)$ in $C_b(X)'$, a bounded sequence $(e_n)$ in E and a sequence $(\alpha _n)\in \ell ^1$ so that

$$\begin{aligned} T(u)=\sum ^\infty _{n=1}\alpha _n \,\Psi _n \,(u)\, e_n \ \ \text{ for } \text{ all } \ \ u\in C_b(X). \end{aligned}$$

(1.1)

Then the space ${\mathcal N}(C_b(X),E)$ of all nuclear operators $T:C_b(X)\rightarrow E$ is a Banach space, equipped with the nuclear norm defined by

$$\begin{aligned} \Vert T\Vert _{nuc}:=\inf \left\{ \sum ^\infty _{n=1}|\alpha _n|\;\Vert \Psi _n\Vert \; \Vert e_n\Vert _E\right\} , \end{aligned}$$

where the infimum is taken over all sequences $(\Psi _n)$ in $C_b(X)'$, $(e_n)$ in E and $(\alpha _n)\in \ell ^1$ such that T admits a representation (1.1) (see [16, Proposition, p. 51]).

2 Nuclear Tight Operators on ${\varvec{C_b(X)}}$

For terminology concerning vector measures, we refer the reader to [5, 6, 18].

Let M(X) denote the Banach lattice of all finitely additive bounded real-valued measures $\nu$ on ${\mathcal B}$ (inner regular by zero-sets), equipped with the norm $\Vert \nu \Vert :=|\nu |(X)$. Due to the Alexandrov representation theorem (see [25, Theorem 5.1]) $C_b(X)'$ can be identified with M(X) through the lattice isomorphism

$$\begin{aligned} \Psi :M(X)\ni \nu \mapsto \Psi _\nu \in C_b(X)', \end{aligned}$$

where $\Psi _\nu (u)=\int _X u(x)\,d\nu$ for all $u\in C_b(X)$ and $\Vert \Psi _\mu \Vert =|\nu |(X)$.

By $B({\mathcal B})$ we denote the Banach lattice (with the uniform norm $\Vert \cdot \Vert _\infty$) of all bounded real functions on X that are uniform limits of sequences of ${\mathcal B}$-simple functions on X. Then $C_b(X)\subset B({\mathcal B})$ (see [1, Lemma 1.2]), and one can embed isometrically $B({\mathcal B})$ in $C_b(X)''$ by the mapping $\pi :B({\mathcal B})\rightarrow C_b(X)''$ , where for $v\in B({\mathcal B})$,

$$\begin{aligned} \pi (v)(\Psi _\nu ):=\int _X v(x)\,d\nu \ \ \text{ for } \text{ all } \ \nu \in M(X). \end{aligned}$$

Let $E'$ and $E''$ denote the dual and the bidual of a Banach space E. Let $i:E\rightarrow E''$ stand for the canonical isometry, that is, $i(e)(e')=e'(e)$ for all $e\in E$ and $e'\in E'$. Let $j:i(E)\rightarrow E$ denote the left inverse of i, that is, $j(i(e))=e$ for $e\in E$.

Assume that $T:C_b(X)\rightarrow E$ is a wekly compact bounded linear operator. Let $T':E'\rightarrow C_b(X)'$ and $T'':C_b(X)''\rightarrow E''$ denote the conjugate and biconjugate of T, respectively. Then by the Gantmacher type theorem (see [2, Theorem 17.2] we have $T''(C_b(X)'')\subset i(E)$. Let us put

$$\begin{aligned} m(A):=j(T''(\pi (\mathbbm {1}_A)))\ \ \text{ for } \text{ all } \ A\in {\mathcal B}. \end{aligned}$$

Then $m:{\mathcal B}\rightarrow E$ is called the representing measure of T and

$$\begin{aligned} T(u)=\int _X u(x)\,dm \ \ \text{ for } \text{ all } \ u\in C_b(X). \end{aligned}$$

The strict topology $\beta$ on $C_b(X)$ has been studied intensively (see [4, 7, 12, 21, 25]). $\beta$ can be characterized as the finest locally convex Hausdorff topology on $C_b(X)$ which coincides with the compact-open topology $\tau _c$ on all $\Vert \cdot \Vert _\infty$-bounded sets in $C_b(X)$ (see [21, Theorem 2.4], [20]). This means that $(C_b(X),\beta )$ is a generalized DF-space (see [20, Corollary]). Then $\beta$ is weaker than the $\Vert \cdot \Vert _\infty$-norm topology on $C_b(X)$, and if, in particular, X is compact, then these topologies coincide.

Due to [7, Lemma 4.5] the dual space $(C_b(X),\beta )'$ of $(C_b(X),\beta )$ can be identified with $rca({\mathcal B}o)$ through the isomorphic isometry

$$\begin{aligned} \Phi :rca({\mathcal B}o)\ni \lambda \mapsto \Phi _\lambda \in (C_b(X),\beta )', \end{aligned}$$

(2.1)

where $\Phi _\lambda (u)=\int _X u(x)\,d\lambda$ for all $u\in C_b(X)$ and $\Vert \Phi _\lambda \Vert = |\lambda |(X)$.

The following characterization of weakly compact tight operators $T:C_b(X)\!\rightarrow \!E$ will be useful (see [14, Theorem 5.5]).

Theorem 2.1

Assume that T is a weakly compact operator and $m:{\mathcal B}\rightarrow E$ is its representing measure. Then the following statements are equivalent:

• (i) T is tight.

• (ii) T is $(\beta ,\Vert \cdot \Vert _E)$-continuous.

• (iii) m can be uniquely extended to a Radon measure ${\widetilde{m}}:{\mathcal B}o\rightarrow E$, that is, ${\widetilde{m}}$ is countably additive and for every $A\in {\mathcal B}o$ and $\varepsilon >0,$ there exist a compact set K and an open set O in X with $K\subset A\subset O$ such that $\Vert {\widetilde{m}}\Vert (O\smallsetminus K)\le \varepsilon$.

Then $T(u)=\int _X u(x)\,dm=\int _X u(x)\,d{\widetilde{m}}$ for all $u\in C_b(X).$

From now on $|{\widetilde{m}}|(A)$ stands for the variation of the measure ${\widetilde{m}}$ on a set $A\in {\mathcal B}o$.

The following result establishes the relationship between nuclear tight operators $T:C_b(X)\rightarrow E$ and their representing measures ${\widetilde{m}}:{\mathcal B}o\rightarrow E$.

Theorem 2.2

Assume that $T:C_b(X)\rightarrow E$ is a weakly compact tight operator and $m:{\mathcal B}\rightarrow E$ is its representing measure. Then the following statements are equivalent:

• (i) T is nuclear.

• (ii) $|{\widetilde{m}}|\in rca({\mathcal B}o)^+$ and the measure ${\widetilde{m}}:{\mathcal B}o\rightarrow E$ has a $|{\widetilde{m}}|$-Bochner integrable derivative.

• (iii) $|{\widetilde{m}}|\in rca ({\mathcal B}o)^+$ and T is Bochner representable with respect to $|{\widetilde{m}}|$.

In this case, $\Vert T\Vert _{nuc}=|{\widetilde{m}}|(X)$.

Proof

(i)$\Leftrightarrow$(ii) See [15, Theorem 3.3].

(ii)$\Rightarrow$(iii) Assume that (ii) holds, that is, there exists a function $f\in L^1(|{\widetilde{m}}|,E)$ such that

$$\begin{aligned} {\widetilde{m}}(A)=\int _A f(x)\,d|{\widetilde{m}}| \ \ \text{ for } \text{ all } \ \ A\in {\mathcal B}o. \end{aligned}$$

Let $u\in C_b(X)$. Since $u\in B({\mathcal B}o)$, there exists a sequence $(s_n)$ of ${\mathcal B}o$-simple functions on X such that $\Vert u-s_n\Vert _\infty \rightarrow 0$. Then for every $n\in {\mathbb {N}}$,

$$\begin{aligned} \int _X s_n(x)\,d{\widetilde{m}}=\int _X s_n(x)\,f(x)\,d|{\widetilde{m}}|, \end{aligned}$$

and hence we have

$$\begin{aligned}&\left\| \int _X u(x)\,f(x)\,d|{\widetilde{m}}|-\int _X s_n(x)\,f(x)\,d|{\widetilde{m}}|\;\right\| _E \le \int _X |u(x)-s_n(x)|\;\Vert f(x)\Vert _E \,d|{\widetilde{m}}|\\&\quad \le \Vert u-s_n\Vert _\infty \int _X \Vert f(x)\Vert _E\, d|{\widetilde{m}}|\rightarrow 0. \end{aligned}$$

Since we have that

$$\begin{aligned} T(u)=\int _X u(x)\, d{\widetilde{m}}=\lim _n\int _X s_n(x)\, d{\widetilde{m}}=\lim _n\int _X s_n(x)\, f(x)\, d|{\widetilde{m}}|, \end{aligned}$$

we get $T(u)=\int _X u(x)\,f(x)\,d|{\widetilde{m}}|$, as desired.

(iii)$\Rightarrow$(ii) This is obvious.

In view of [15, Theorem 3.3] we have that $\Vert T\Vert _{nuc}=|{\widetilde{m}}|(X)$. $\square$

Grothendieck carried over the concept of nuclear operators to locally convex spaces (see [26, p. 289], [19, Chap. 3, Sect. 7], [13, Sect. 17.3], [24, Sect. 47]).

We will need the following result (see [21, Theorem 5.1]).

Theorem 2.3

For a subset ${\mathcal M}$ of $rca({\mathcal B}o)$ the following statements are equivalent:

• (i) $\{\Phi _\lambda :\lambda \in {\mathcal M}\}$ is $\beta$-equicontinuous.

• (ii) $\sup _{\lambda \in {\mathcal M}}\Vert \lambda \Vert <\infty$ and ${\mathcal M}$ is uniformly tight, i.e., for every $\varepsilon >0,$ there exists a compact set K in X such that $\sup _{\lambda \in {\mathcal M}}|\lambda |(X\smallsetminus K)\le \varepsilon$.

Following [19, Chap. 3, Sect. 7] (see also [24, Sect. 47], [13, 17.3, p. 379]) and using Theorem 2.3 we have the following definition.

Definition 2.1

A linear operator $T:C_b(X)\rightarrow E$ is said to be a nuclear operator between the locally convex space $(C_b(X),\beta )$ and a Banach space E, if there exist a bounded uniformly tight sequence $(\lambda _n)$ in $rca({\mathcal B}o)$, a bounded sequence $(e_n)$ in E and a sequence $(\alpha _n)\in \ell ^1$ so that

$$\begin{aligned} T(u)=\sum ^\infty _{n=1}\alpha _n\left( \int _X u(x)\,d\lambda _n\right) e_n \ \ \text{ for } \text{ all } \ u\in C_b(X). \end{aligned}$$

(2.2)

Then T is $(\beta ,\Vert \cdot \Vert _E)$-compact, that is, T(V) is relatively norm compact in E for some $\beta$-neighborhood V of zero in $C_b(X)$ (see [19, Chap. 3, Sect. 7, Corollary 1]). Hence T is $(\beta ,\Vert \cdot \Vert _E)$-continuous. Let us put

$$\begin{aligned} \Vert T\Vert _{\beta -nuc}:=\inf \left\{ \sum ^\infty _{n=1} |\alpha _n|\;|\lambda _n|(X)\,\Vert e_n\Vert _E\right\} , \end{aligned}$$

where the infimum is taken over all sequences $(\lambda _n)$ in $rca({\mathcal B}o)$, $(e_n)$ in E and sequences $(\alpha _n)\in \ell ^1$ such that T admits a representation (2.2).

Now we can state the following result.

Corollary 2.4

For a linear subset operator $T:C_b(X)\rightarrow E,$ the following statements are equivalent:

• (i) T is a nuclear tight operator.

• (ii) T is a nuclear operator between the locally convex space $(C_b(X),\beta )$ and a Banach space E.

In this case, $\Vert T\Vert _{nuc}=\Vert T\Vert _{\beta -nuc}$.

Proof

(i)$\Rightarrow$(ii) Assume that (i) holds. Then in view of Theorem 2.2$|{\widetilde{m}}|\in rca({\mathcal B}o)^+$ and there exists a function $f\in L^1(|{\widetilde{m}}|,E)$ so that

$$\begin{aligned} T(u)=\int _X u(x)\,f(x)\,d|{\widetilde{m}}| \ \ \text{ for } \text{ all } \ \ u\in C_b(X) \end{aligned}$$

and

$$\begin{aligned} \Vert T\Vert _{nuc}=|{\widetilde{m}}|(X)=\int _X \Vert f(x)\Vert _E \,d|{\widetilde{m}}|. \end{aligned}$$

Let $L^1(|{\widetilde{m}}|){\hat{\otimes }}_\gamma E$ denote the projective tensor product of $L^1(|{\widetilde{m}}|)$ and E, equipped with the completed norm $\gamma$ (see [5, p. 227], [18, p. 17]). Note that for $w\in$ $L^1(|{\widetilde{m}}|)\,{\hat{\otimes }}_\gamma \,E$, we have

$$\begin{aligned} \gamma (w)=\inf \left\{ \sum ^\infty _{n=1}|\alpha _n|\;\Vert v_n\Vert _1\;\Vert e_n\Vert _E\right\} , \end{aligned}$$

where the infimum is taken over all sequences $(v_n)$ in $L^1(|{\widetilde{m}}|)$ and $(e_n)$ in E with $\lim _n\Vert v_n\Vert _1=0=\lim _n\Vert e_n\Vert _E$ and $(\alpha _n)\in \ell ^1$ such that $w=\sum ^\infty _{n=1} \alpha _n v_n\otimes \,e_n$ in $\gamma$-norm (see [18, Proposition 2.8, pp. 21–22]).

It is known that $L^1(|{\widetilde{m}}|){\hat{\otimes }}_\gamma E$ is isometrically isomorphic to the Banach space $(L^1(|{\widetilde{m}}|,E),\Vert \cdot \Vert _1)$ by the isometry $J:L^1(|{\widetilde{m}}|)\,{\hat{\otimes }}_\gamma \,E\rightarrow L^1(|{\widetilde{m}}|,E)$, defined by

$$\begin{aligned} J(v\otimes e):=v(\cdot )\, e \ \ \text{ for } \text{ all } \ \ v\in L^1(|{\widetilde{m}}|), \ e\in E \end{aligned}$$

(see [5, Example 10, p. 228], [18, Example 2.19, p. 29]).

Let $\varepsilon >0$ be given. Then there exist sequences $(v_n)$ in $L^1(|{\widetilde{m}}|)$ and $(e_n)$ in E with $\lim _n\Vert v_n\Vert _1=0=\lim _n\Vert e_n\Vert _E$ and $(\alpha _n)\in \ell ^1$ so that

$$\begin{aligned} J^{-1}(f)=\sum ^\infty _{n=1}\alpha _n v_n\otimes e_n \ \ \text{ in } \text{ the } \gamma \text{-norm } \end{aligned}$$

and

$$\begin{aligned} \sum ^\infty _{n=1}|\alpha _n|\,\;\Vert v_n\Vert _1\;\Vert e_n\Vert _E\le \gamma (J^{-1}(f))+\varepsilon = \Vert f\Vert _1+\varepsilon =|{\widetilde{m}}|(X)+\varepsilon . \end{aligned}$$

(2.3)

Thus this follows that

$$\begin{aligned} f=J\left( \sum ^\infty _{n=1} \alpha _n v_n\otimes e_n\right) =\sum ^\infty _{n=1} \alpha _n v_n(\cdot )\, e_n \end{aligned}$$

and hence

$$\begin{aligned} T(u)=\sum ^\infty _{n=1}\alpha _n\left( \int _X u(x)\, v_n(x)\, d|{\widetilde{m}}|\right) e_n \ \ \text{ for } \text{ all } \ \ u\in C_b(X). \end{aligned}$$

For $n\in {\mathbb {N}}$, let $\lambda _n(A):=\int _A v_n(x)\,d|{\widetilde{m}}|$ for all $A\in {\mathcal B}o$. Then we have

$$\begin{aligned} \Phi _{\lambda _n}(u)=\int _X u(x)\,d\lambda _n=\int _X u(x)\,v_n(x)\,d|{\widetilde{m}}| \ \ \text{ for } \text{ all } \ \ u\in C_b(X) \end{aligned}$$

(see [3, Theorem 8C, p. 380]). Note that $\lambda _n\in rca({\mathcal B}o)$ and $|\lambda _n|(X)=\Vert v_n\Vert _1$. We shall show that the family $\{\lambda _n:n\in {\mathbb {N}}\}$ is uniformly tight.

Indeed, let $\varepsilon >0$ be given. Since $\lim \Vert v_n\Vert _1=0$, we can choose $n_\varepsilon \in {\mathbb {N}}$ such that $|\lambda _n|(X)\le \varepsilon$ for $n>n_\varepsilon$. For $n=1,\dots ,n_\varepsilon$, choose a compact set $K_n$ in X such that $|\lambda _n|(X\smallsetminus K_n)\le \varepsilon$. Denote $K=\bigcup ^{n_\varepsilon }_{n=1}K_n$. Then $|\lambda _n|(X\smallsetminus K)\le \varepsilon$ for all $n\in {\mathbb {N}}$, as desired.

Since $T(u)=\sum ^\infty _{n=1}\alpha _n(\int _X u(x)\,d\lambda _n)e_n$ for all $u\in C_b(X)$, we see that T is a nuclear operator between $(C_b(X),\beta )$ and a Banach space E, and in view of (2.3) we get

$$\begin{aligned} \Vert T\Vert _{\beta -nuc}\le |{\widetilde{m}}|(X)=\Vert T\Vert _{nuc}. \end{aligned}$$

Since $\Vert T\Vert _{nuc}\le \Vert T\Vert _{\beta -nuc}$, we have that $\Vert T\Vert _{nuc}=\Vert T\Vert _{\beta -nuc}$.

(ii)$\Rightarrow$(i) Assume that (ii) holds. Then T is nuclear and $(\beta ,\Vert \cdot \Vert _E)$-continuous. Hence by Theorem 2.1T is tight. $\square$

As a consequence of Corollary 2.4, we get

Corollary 2.5

Assume that $T:C_b(X)\rightarrow E$ is a nuclear tight operator and $m:{\mathcal B}\rightarrow E$ is its representing measure. Then the mapping

$$\begin{aligned} T^*:E'\ni e'\mapsto e'\circ {\widetilde{m}}\in rca({\mathcal B}o) \end{aligned}$$

is a nuclear operator and $\Vert T^*\Vert _{nuc}=\Vert T\Vert _{nuc}=|{\widetilde{m}}|(X).$

Proof

Let $\varepsilon >0$ be given. According to Corollary 2.4, there exist bounded sequences $(\lambda _n)$ in $rca({\mathcal B}o)$ and $(e_n)$ in E and a sequence $(\alpha _n)\in \ell ^1$ so that

$$\begin{aligned} T(u)=\sum ^\infty _{n=1}\alpha _n\Phi _{\lambda _n}(u)\, e_n \ \ \text{ for } \text{ all } \ \ u\in C_b(X) \end{aligned}$$

and

$$\begin{aligned} \sum ^\infty _{n=1}|\alpha _n|\;|\lambda _n|(X)\,\Vert e_n\Vert _E\le |{\widetilde{m}}|(X)+\varepsilon . \end{aligned}$$

(2.4)

Thus it follows that for each $e'\in E'$, we have

$$\begin{aligned} e'\circ T=\sum ^\infty _{n=1}\alpha _n\,e'(e_n)\,\Phi _{\lambda _n} \ \ \text{ in } \ \ C_b(X)'. \end{aligned}$$

By Theorem 2.1 for each $e'\in E'$, we have

$$\begin{aligned} (e'\circ T)(u)=\int _X u(x)\,d\,e'\circ {\widetilde{m}}\ \ \text{ for } \text{ all } \ \ u\in C_b(X), \end{aligned}$$

where $e'\circ {\widetilde{m}}\in rca({\mathcal B}o)$. Hence we have

$$\begin{aligned} T^*(e'):=e'\circ {\widetilde{m}}=\Phi ^{-1}(e'\circ T)=\sum ^\infty _{n=1}\alpha _n \,i(e_n)\,(e')\,\lambda _n. \end{aligned}$$

This means that $T^*$ is a nuclear operator and in view of (2.4) we get

$$\begin{aligned} \Vert T^*\Vert _{nuc}\le |{\widetilde{m}}|(X). \end{aligned}$$

Now we shall show that

$$\begin{aligned} |{\widetilde{m}}|(X)\le \Vert T^*\Vert _{nuc}. \end{aligned}$$

Indeed, let $\varepsilon >0$ be given. Since $T^*$ is a nuclear operator, there exist a bounded sequence $(e''_n)$ in $E''$, a bounded sequence $(\lambda _n)$ in $rca({\mathcal B}o)$ and $(\alpha _n)\in \ell ^1$ so that

$$\begin{aligned} T^*(e')=\sum ^\infty _{n=1} \alpha _n \,e''_n(e')\,\lambda _n \ \ \text{ for } \ \ e'\in E' \end{aligned}$$

and

$$\begin{aligned} \sum ^\infty _{n=1}|\alpha _n|\;\Vert e''_n\Vert _{E''}\;|\lambda _n|(X)\le \Vert T^*\Vert _{nuc}+\varepsilon . \end{aligned}$$

(2.5)

Then for $A\in {\mathcal B}o$, we have

$$\begin{aligned} (e'\circ {\widetilde{m}})(A)=T^*(e')(A)=\sum ^\infty _{n=1}\alpha _n \,e''_n(e')\,\lambda _n(A). \end{aligned}$$

By the Hahn-Banach theorem for every $A\in {\mathcal B}o$, there exists $e'_A\in E'$ with $\Vert e'_A\Vert _{E'}=1$ such that $\Vert {\widetilde{m}}(A)\Vert _E=|(e'_A\circ {\widetilde{m}})(A)|$. Hence, if $\Pi$ is a finite ${\mathcal B}o$-partition of X, then using (2.5) we get

$$\begin{aligned} \displaystyle \sum _{A\in \Pi }\Vert {\widetilde{m}}(A)\Vert _E= & {} \sum _{A\in \Pi }|(e'_A\circ {\widetilde{m}})(A)|=\sum _{A\in \Pi } \left| \sum ^\infty _{n=1} \alpha _n\, e''_n(e'_A)\,\lambda _n(A)\right| \\\le & {} \sum _{A\in \Pi }\left( \sum ^\infty _{n=1} |\alpha _n|\;|e''_n(e'_A)|\; |\lambda _n(A)|\right) \le \sum ^\infty _{n=1}\left( |\alpha _n|\;\Vert e''_n\Vert _E \sum _{A\in \Pi }|\lambda _n(A)|\right) \\\le & {} \sum ^\infty _{n=1}|\alpha _n|\;\Vert e''_n\Vert _{E''}\, |\lambda _n|(X)\le \Vert T^*\Vert _{nuc}+\varepsilon . \end{aligned}$$

Since $\varepsilon >0$ is arbitrary, we get $|{\widetilde{m}}|(X)\le \Vert T^*\Vert _{nuc}$ and hence $\Vert T^*\Vert _{nuc}=|{\widetilde{m}}|(X)=\Vert T\Vert _{nuc}$. $\square$

3 Bochner Representable Operators on ${C_b(X)}$

Making use of [6, Sect. 2, F, Theorem 30, p. 32], we have the following

Lemma 3.1

For $\mu \in rca({\mathcal B}o)^+$ and $g\in L^1(\mu ,E)$, let us put

$$\begin{aligned} \lambda (A):=\int _A\Vert g(x)\Vert _E\,d\mu \ \ \text{ for } \text{ all } \ \ A\in {\mathcal B}o \end{aligned}$$

and $f_g(x):=g(x)/\Vert g(x)\Vert _E$ if $g(x)\ne 0$ and $f_g(x):=0$ if $g(x)=0.$

Then

$$\begin{aligned} f_g\in L^1(\lambda ,E) \ \ \text{ and } \ \int _X v(x)\,f_g(x)\,d\lambda =\int _X v(x)\,g(x) \,d\mu \ \ \text{ for } \text{ all } \ \ v\in B({\mathcal B}o). \end{aligned}$$

In particular, $\int _A f_g(x)\,d\lambda =\int _A g(x)\,d\mu$ for all $A\in {\mathcal B}o.$

Now we can characterize Bochner representable operators $T:C_b(X)\rightarrow E$.

Theorem 3.2

If $T:C_b(X)\rightarrow E$ is a Bochner representable operator, then T is a nuclear tight operator.

Proof

There exist $\mu \in rca({\mathcal B}o)^+$ and $g\in L^1(\mu ,E)$ so that

$$\begin{aligned} T(u)=\int _X u(x)\,g(x)\,d\mu \ \ \text{ for } \text{ all } \ \ u\in C_b(X). \end{aligned}$$

Define

$$\begin{aligned} S(v):=\int _X v(x)\,g(x)\,d\mu \ \ \text{ for } \text{ all } \ \ v\in B({\mathcal B}o) \end{aligned}$$

and let

$$\begin{aligned} m_S(A):=S(\mathbbm {1}_A)=\int _A g(x)\,d\mu \ \ \text{ for } \text{ all } \ \ A\in {\mathcal B}o. \end{aligned}$$

Then

$$\begin{aligned} S(v)=\int _X v(x)\,dm_S \ \ \text{ for } \text{ all } \ \ v\in B({\mathcal B}o) \end{aligned}$$

and

$$\begin{aligned} |m_S|(A)=\int _A\Vert g(x)\Vert _E\,d\mu \ \ \text{ for } \text{ all } \ \ A\in {\mathcal B}o. \end{aligned}$$

(see [5, Theorem 13, p. 6 and Theorem 4, p. 46]). Since $|m_S|$ is a $\mu$-absolutely continuous measure, we obtain that $|m_S|\in rca({\mathcal B}o)^+$ and hence $m_S:{\mathcal B}o\rightarrow E$ is a Radon measure. Thus, it follows that $S:B({\mathcal B}o)\rightarrow E$ is weakly compact (see [5, Theorem 1, p. 148]), and hence $T:C_b(X)\rightarrow E$ is weakly compact and

$$\begin{aligned} T(u)=\int _X u(x)\,dm_S \ \ \text{ for } \text{ all } \ \ u\in C_b(X). \end{aligned}$$

(3.1)

To show that T is a tight operator, assume that $(u_\alpha )$ is a net in $C_b(X)$ such that $\sup _\alpha \Vert u_\alpha \Vert _\infty =a<\infty$ and $u_\alpha \rightarrow 0$ uniformly on all compact sets in X.

Let $\varepsilon >0$ be given and choose $\delta >0$ such that

$$\begin{aligned} \int _A\Vert g(x)\Vert _E \,d\mu \le \frac{\varepsilon }{2a} \ \ \text{ for } \text{ all } \ \ A\in {\mathcal B}o \ \ \text{ with } \ \ \mu (A)\le \delta . \end{aligned}$$

Choose a compact subset $K_\varepsilon$ of X such that $\mu (X\smallsetminus K_\varepsilon ) \le \delta$. Then there exists $\alpha _0$ such that

$$\begin{aligned} \sup _{x\in K_\varepsilon }|u_\alpha (x)|\le \frac{\varepsilon }{2|m_S|(X)} \ \ \text{ for } \ \ \alpha \ge \alpha _0. \end{aligned}$$

Hence for $\alpha \ge \alpha _0$, we get

$$\begin{aligned} \Vert T(u_\alpha )\Vert _E\le & {} \int _X |u_\alpha (x)|\;\Vert g(x)\Vert _E\,d\mu \\= & {} \int _{K_\varepsilon }|u_\alpha (x)|\;\Vert g(x)\Vert _E\,d\mu +\int _{X\smallsetminus K_\varepsilon } |u_\alpha (x)|\;\Vert g(x)\Vert _E\,d\mu \\\le & {} \frac{\varepsilon }{2|m_S|(X)}\int _X\Vert g(x)\Vert _E\,d\mu +\Vert u_\alpha \Vert _\infty \int _{X\smallsetminus K_\varepsilon }\Vert g(x)\Vert _E\,d\mu \le \frac{\varepsilon }{2}+\frac{\varepsilon }{2}=\varepsilon . \end{aligned}$$

This means that T is a tight operator, as desired. Then by Theorem 2.1 and (3.1) we get

$$\begin{aligned} \int _X u(x)\,d{\widetilde{m}}=\int _X u(x)\,dm_S \ \ \text{ for } \text{ all } \ \ u\in C_b(X), \end{aligned}$$

and hence for each $e'\in E'$, we have

$$\begin{aligned} \int _X u(x)\,d\,e'\circ {\widetilde{m}}=\int _X u(x)\,d\, e'\circ m_S \ \ \text{ for } \text{ all } \ \ u\in C_b(X). \end{aligned}$$

Since $e'\circ \,{\widetilde{m}}\in rca({\mathcal B}o)$ and $e'\circ m_S\in rca({\mathcal B}o)$, in view of (2.1) we get $e'\circ \,{\widetilde{m}}=e'\circ \,m_S$ for every $e'\in E'$. Thus this follows that ${\widetilde{m}}=m_S$, that is, for all $A\in {\mathcal B}o$,

$$\begin{aligned} {\widetilde{m}}(A)=\int _A g(x)\,d\mu \ \ \text{ and } \ \ |{\widetilde{m}}|(A)=\int _A\Vert g(x)\Vert _E\,d\mu . \end{aligned}$$

By Lemma 3.1$f_g\in L^1(|{\widetilde{m}}|,E)$ and for all $A\in {\mathcal B}o$,

$$\begin{aligned} {\widetilde{m}}(A)=\int _A g(x)\,d\mu =\int _A f_g(x)\,d|{\widetilde{m}}|. \end{aligned}$$

According to Theorem 2.2 this means that T is a nuclear operator. $\square$

Now we can state an extension of Tong’s result (see [23, Theorem 1.2]) to the setting of completely regular Hausdorff spaces.

Corollary 3.3

For a linear operator $T:C_b(X)\rightarrow E$, the following statements are equivalent:

• (i) T is nuclear and tight.

• (ii) T is Bochner representable.

• (iii) T is a nuclear operator between the locally convex space $(C_b(X),\beta )$ and a Banach space E.

Proof

(i)$\Rightarrow$(ii) This follows from Theorem 2.2.

(ii)$\Rightarrow$(i) This follows from Theorem 3.2.

(i)$\Leftrightarrow$(iii) See Corollary 2.4. $\square$